Problem: Divide the following complex numbers. $ \dfrac{19+25i}{2-5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${2+5i}$ $ \dfrac{19+25i}{2-5i} = \dfrac{19+25i}{2-5i} \cdot \dfrac{{2+5i}}{{2+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(19+25i) \cdot (2+5i)} {(2-5i) \cdot (2+5i)} = \dfrac{(19+25i) \cdot (2+5i)} {2^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(19+25i) \cdot (2+5i)} {(2)^2 - (-5i)^2} = $ $ \dfrac{(19+25i) \cdot (2+5i)} {4 + 25} = $ $ \dfrac{(19+25i) \cdot (2+5i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({19+25i}) \cdot ({2+5i})} {29} = $ $ \dfrac{{19} \cdot {2} + {25} \cdot {2 i} + {19} \cdot {5 i} + {25} \cdot {5 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{38 + 50i + 95i + 125 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{38 + 50i + 95i - 125} {29} = \dfrac{-87 + 145i} {29} = -3+5i $